Symmetry properties and exact solutions of the time fractional
Kolmogorov-Petrovskii-Piskunov equation
M.S. Hashemia, M. Incb, and M. Bayramc aDepartment of Mathematics, Basic Science Faculty,
University of Bonab, Bonab 55517, Iran. bDepartment of Mathematics, Science Faculty,
Firat University, 23119 Elazig, Turkey. cDepartment of Computer Engineering, Istanbul Gelisim University, Istanbul, Turkey. Received 17 January 2019; accepted 19 February 2019
In this paper, the time fractional Kolmogorov-Petrovskii-Piskunov (TFKPP) equation is analyzed by means of Lie symmetry approach. The TFKPP is reduced to ordinary differential equation of fractional order via the attained point symmetries. Moreover, the simplest equation method is used in construct the exact solutions of underlying equation with recently introduced conformable fractional derivative.
Keywords: Time fractional Kolmogorov-Petrovskii-Piskunov equation; Lie symmetry analysis; Erd´elyi-Kober fractional derivative; Riemann-Liouville derivative; conformable fractional derivative; simplest equation method.
PACS: 02.30.Uu; 04.20.Jb; 05.40.Fb; 05.60.-k DOI: https://doi.org/10.31349/RevMexFis.65.529
1.
Introduction
The fractional calculus (FC) began to wind up exceptionally famous in a few parts of science and engineering. Numer-ous important event, that is, acNumer-oustics, anomalNumer-ous diffusion, chemistry, control processing, electro-magnetics, and visco-elasticity have been expressed by FC. It is known that a sys-tematic method for extracting the analytical solution of both ordinary differential equations (ODEs) and partial differen-tial equations (PDEs) was first proposed by the Norwegian mathematician Sophus Lie in the early 19th century. The fun-damental overview of this strategy is the estimation of vari-able changes that can leave differential condition unchanged. Therefore, a vital role in the field of FC is to attain the Lie symmetries and the solutions of the equations with the FC derivatives. There have been some properties of the frac-tional sense that could not be found in classical sense, ow-ing to this we feel motivated to establish the symmetries of TFKPP equation. This equation has the following general-ized form [1-4]
∂α
tu = uxx+ λu + µu2+ γu3, λ + µ + γ = 0,
ψ2= µ2− 4λγ ≥ 0, (1)
where ∂tαu := Dα
tu stands for Riemann-Liouville of order
α, expressed as [5] Dtαu(x, t) = 1 Γ(n−α) ∂ n ∂tn t R 0 (t − ξ)n−α−1u(x, ξ)dξ, n − 1 < α < n ∂nu ∂tn, α = n ∈ N (2) The TFKPP Eq. (1), has a large application and includes as particular cases the time fractional Fitzhugh-Nagumo
equation (λ = −c, µ = c + 1, γ = −1, 0 < c < 1), which is used in population genetics, the time fractional Newell-Whitehead equation (λ = 1, µ = 0, γ = −1). Recently, the homotopy perturbation method and homotopy analysis method have utilized to consider the TFKPP equation by Gepreel [1] and Hariharan [2], respectly with λ = µ = 0 and γ = −2.
In FC, there are large amount of differential derivatives were defined e.g. [6-9]. In the calculus, the chain rule is a useful and an applicable. It is also hold for conformable frac-tional derivatives.
As far as we know, every proposed fractional derivative has some disadvantages. Therefore, Khalil et al., [9], pro-posed a new definitions:
Definition 1.1. Surmise that f : [a, b] × (0, ∞) → R, then the conformable fractional derivative of f is given by
tTα(f )(x, t) = lim ²→0 f¡x, t + ²t1−α¢− f (x, t) ² , α ∈ (0, 1], (3) for all t > 0.
Theorem 1.1 [9] Suppose that a, b ∈ R and α ∈ (0, 1], then
(i)tTα(au + bv) = atTα(u) + btTα(v), (ii)tTα(tλ) = λtλ−α, λ ∈ R,
(iii)tTα(uv) = utTα(v) + vtTα(u), (iv)tTα ³ u v ´ = utTα(v)−vtTα(u) v2 , (v)tTα(u)(t) = t1−αu0(t), u ∈ C1.
More than that, the chain rule is valid for conformable fractional derivatives, shown by Abdeljawad [10].
Theorem 1.2. Surmise that f : (0, ∞) → R is a real differentiable, α−differentiable function. Assume that g is a function defined in the range of f and also differentiable; then, one has the following rule:
tTα(f og)(t) = t1−αg0(t)f0(g(t)). (4) There are many investigation about conformable frac-tional derivatives [11-14] and also some physical interpre-tations of this newly introduced fractional derivative are de-scribed in [15].
The organization of the manuscript is given below: In Sec. 2, we provide some preliminaries. Section 3, is de-voted to the description of Lie symmetry analysis of TFKPP Eq. (1). General similarity forms and symmetry reductions are established. In Sec. 4, exact solutions to the TFKPP equa-tion with conformable fracequa-tional derivative are investigated. Finally, the last section is devoted to conclusions.
2.
Lie symmetry analysis of fractional partial
differential equations
Here, some description for solving fractional partial differ-ential equations (FPDEs) via Lie symmetry analysis will be provided. Surmise that FPDE having as in [16-26]
∂α
tu = F (x, t, u, ux, uxx), 0 < α < 1. (5) If (5) is invariant under a one parameter Lie group of point transformations
¯t = ¯t(x, t, u; ²), ¯x = ¯x(x, t, u; ²), ¯u = ¯u(x, t, u; ²), (6)
the vector field of an evolution type of equation is as follows:
V = ξt(x, t, u)∂ ∂t+ ξ x(x, t, u) ∂ ∂x + φ(x, t, u) ∂ ∂u, (7)
where the coefficients ξt, ξxand φ of the vector field are to be determined. When V satisfy the Lie symmetry condition, the vector field (7) generates a symmetry of (5),
pr(α,2)V (∆)|∆=0= 0, ∆ = ∂tαu − F. Thus the extension operator take the form
pr(α,2)V = V + φ0 α∂∂α tu+ φ x∂ ux+ φ xx∂ uxx, where φx= D x(φ) − uxDx(ξx) − utDx(ξt), φxx= D x(φx) − uxtDx(ξt) − uxxDx(ξx), φ0 α= Dαt(φ) + ξxDαt(ux) − Dtα(ξxux) + Dα t(Dt(ξt)u) − Dα+1t (ξtu) + ξtDα+1t (u). The condition of invariance
ξt(x, t, u)|
t=0= 0,
is inevitable for the (6), due to the (2).
The αthextended infinitesimal is presented as:
φ0
α= Dαt(φ) + ξxDαt(ux) − Dtα(ξxux) + Dαt(Dt(ξt)u)
− Dα+1
t (ξtu) + ξtDtα+1(u), (8) where Dtα exhibits the total fractional derivative operator. The fractional generalized Leibnitz rule is expressed as
Dα t £ u(t)v(t)¤= ∞ X n=0 µ α n ¶ Dα−n t u(t)Dntv(t), α > 0, (9) here µ α n ¶ =(−1) n−1αΓ(n − α) Γ(1 − α)Γ(n + 1) .
Therefore using (9) one can represent (8) as
φ0 α= Dαt(φ) − αDt(ξt)∂ αu ∂tα − ∞ X n=1 µ α n ¶ Dn t(ξx)Dα−nt (ux) − ∞ X n=1 µ α n + 1 ¶ Dn+1 t (ξt)Dα−nt (u). (10)
Using chain rule
dmf (g(t)) dtm = m X k=0 k X r=0 µ k r ¶ × 1 k![−g(t)] rdm dtm[g(t) k−r]dkf (g) dgk , and setting f (t) = 1, one can get
Dα t(φ) = ∂αφ ∂tα + φu ∂αu ∂tα − u ∂αφ u ∂tα + ∞ X n=1 µ α n ¶ ∂nφ u ∂tn D α−n t (u) + ϑ, where ϑ = ∞ X n=2 n X m=2 m X k=2 k−1X r=0 µ α n ¶ µ n m ¶ µ k r ¶ 1 k! × t n−α Γ(n + 1 − α)[−u] r∂m ∂tm[u k−r]∂n−m+kφ ∂tn−m∂uk. Therefore φ0 α= ∂αφ ∂tα + ¡ φu− αDt(ξt)¢ ∂ αu ∂tα − u ∂αφ u ∂tα + ϑ + ∞ X n=1 £µ α n ¶ ∂αφ u ∂tα − µ α n + 1 ¶ Dn+1 t (ξt) ¤ Dα−n t (u) − ∞ X n=1 µ α n ¶ Dn t(ξx)Dα−nt (ux).
3.
Symmetry representation of TFKPP
equa-tion
In view of the Lie theory, we have:
φ0
α= φxx+ λφ + 2µφu + 3γφu2. (11) Substituting (10) into (11), the determining equations for Eq. (1) is attained,consequently, we have
ξt= 4tc
3, ξx= c1+ 2αxc3,
φ = c2u + (3α − 2)uc3+ C(x, t),
where c1, c2and c3are constants and C(x, t) is a solution of
Eq. (1). Therefore, the algebra g of Eq. (1) can be written as
V1= ∂ ∂x, V2= u ∂ ∂u, V3= 4t∂ ∂t + 2αx ∂ ∂x+ (3α − 2)u ∂ ∂u, V4= C(x, t) ∂ ∂u.
For V3, one can write
dt 4t = dx 2αx = du (3α − 2)u,
and this give
ζ = xt−α2 , u(x, t) = t3α−24 F(ζ). (12)
Theorem 3.1. The transformation (12) reduces (1) to the following: ³ P−α4+12,α 2 α F ´ (ζ) = F00+ λF + µF2+ γF3, (13)
with the Erd´elyi-Kober (EK) fractional differential operator
Pβτ,αdefined by ³ Pβτ,αF´: = n−1Y j=0 µ τ + j − 1 βζ d dζ ¶ ³ Kτ +α,n−αβ F´(ζ), n = ½ [α] + 1, α /∈ N α, α ∈ N where ³ Kτ,αβ F´(ζ) := 1 Γ(α) ∞ R 1 (u − 1)α−1u−(τ +α)F(ζuβ1)du, F(ζ), α = 0,
is the EK fractional integral operator.
Proof: Let n − 1 < α < n, n = 1, 2, 3, .... By means of Reimann-Liouville, one reaches
∂αu ∂tα = ∂n ∂tn · 1 Γ(n − α) × t Z 0 (t − s)n−α−1s3α−24 F ³ xs−α2 ´ ds ¸ . (14)
Letting % = t/s, one can get ds = −(t/%2)d%, therefore (14) can be written as ∂αu ∂tα = ∂n ∂tn · tn−α4−12 ³ K3α+24 ,n−α 2 α F ´ (ζ) ¸ .
Taking into account the relation (ζ = xt−α/2), we can obtain
t∂ ∂tφ(ζ) = t ∂ζ ∂t dφ(ζ) dζ = − α 2ζ dφ(ζ) dζ .
Therefore one can get
∂n ∂tn · tn−α 4−12 ³ K3α+24 ,n−α 2 α F ´ (ζ) ¸ = ∂n−1 ∂tn−1 · ∂ ∂t ³ tn−α 4−12 ³ K3α+24 ,n−α 2 α F ´ (ζ)´ ¸ = ∂n−1 ∂tn−1 · tn−α 4−32 µ n −α 4 − 1 2 − α 2ζ d dζ ¶ × ³ K3α+24 ,n−α 2 α F ´ (ζ) ¸ = . . . = t−α 4−12 n−1Y j=0 µ −α 4 + 1 2+ j − α 2ζ d dζ ¶ ׳K3α+24 ,n−α 2 α F ´ (ζ) = t−α 4−12 ³ P−α4+12,α 2 α F ´ (ζ).
This completes the proof.
Also, for the symmetry of V1+ V2+ V3, one can write
dt 4t = dx 2αx + 1 = du (3α − 1)u, which yields ζ = 2αx + 1 2α t −α 2 , u(x, t) = t3α−14 F(ζ). (15)
Theorem 3.2. The transformation (15) reduces (1) to the following nonlinear ordinary differential equation of frac-tional order: ³ P3−α4 ,α 2 α F ´ (ζ) = F00+ λF + µF2+ γF3. (16)
4.
Exact Solutions of TFKPP equation
Symmetry analysis of differential equations gives many in-formation about geometric properties of various differential equations. For example, it is possible to extract vector fields, infinitesimals, conservation laws and reductions of differ-ential equations. Reduction procedure of differdiffer-ential equa-tions allows us to reduce dimension of these equaequa-tions by one less. In two dimensional partial differential equations (PDEs), reduction procedure gives an ordinary differential equation (ODE). So, solving this ODE concludes exact so-lution of original PDE. However, in FPDEs with Riemann-Liouville fractional derivatives we get ODEs with the EK derivatives which there is not a systematic method to find their exact solution. Therefore, after reduction of TFKPP equation with the Riemann-Liouville fractional derivative we obtain Eqs. (13) and (16) which it is not possible to find an-alytical solutions. However, we can obtain exact solution of Eq. (1) with ∂tαu := tTα(u). In this section, we investi-gate the exact solutions of TFKPP equation with conformable fractional derivative.
4.1. Simplest equation method and its applications to time fractional differential equations
This approach was proposed in [27,28]. The steps for the approach is stated as follows:
Let the TFDE is given by
P (u, tTα(u), ux, uxx, ...) = 0. 0 < α ≤ 1, (17) Then the modified version of simplest equation method pro-cedure have the following steps:
Step 1: We utilize the following
u(x, t) = Θ(ξ), ξ = A µ x − νt α α ¶ , (18)
where A and ν are nonzero constants to be determined later. Consequently we attain with parameters A and ν the fol-lowing
P¡Θ, −AνΘ0, AΘ0, A2Θ00, ...¢= 0. (19) Step 2: Suppose that Eq. (19) possesses
Θ(ξ) = N X i=0
ai[z(ξ)]i, (20)
where ai, i = 0, 1, ..., N , are constants to be determined later. The positive value of N in (20), which the pole order for the general solution of Eq. (19), can be determined by substituting Θ(ξ) = ξ−m, (m > 0).
In the present paper, we use the Bernoulli and Riccati equations which their solutions can be expressed by elemen-tary functions. For the Bernoulli equation:i
dz
dξ = az(ξ) + b[z(ξ)]
k, k ∈ N\{1},
we use the solutions
z(ξ) = k−1 s a exp£a(k − 1)(ξ + ξ0) ¤ 1 − b exp£a(k − 1)(ξ + ξ0) ¤ ,
for the case a > 0, b < 0 and
z(ξ) = k−1 s − a exp £ a(k − 1)(ξ + ξ0) ¤ 1 + b exp£a(k − 1)(ξ + ξ0) ¤ ,
for the case a < 0, b > 0 and ξ0is a constant of integration.
For the Riccati equation
dz
dξ = a + b[z(ξ)]
2,
which admits the following exact solutions:
z(ξ) = − √ −ab b tanh ·√ −abξ−² ln(ξ0) 2 ¸ , ξ0> 0, ² = ±1, when ab < 0 and z(ξ) = √ ab b tan £√ abξ + ξ0 ¤ , ξ0= Const., when ab > 0.
Step 3: Plugging (20) into (19) and equating the coef-ficients of zito zero, one can obtain an algebraic system in
A, ν and ai, i = 0, ..., N .
4.2. Application to the TFKPP equation The transformation u(x, t) = Θ(ξ), ξ = A µ x − νtα α ¶ , (21)
changes Eq. (1) with ∂tαu = tTα(u) to:
A2Θ00+ νAΘ0+ λΘ + µΘ2+ γΘ3= 0. (22) We suppose that Eq. (22) has solution of the form (20). Balancing the highest order derivative terms with nonlinear terms in Eq. (22), we get N = 1, and hence
Θ(ξ) = a0+ a1z(ξ), a16= 0. (23)
Substituting (23) along with (21) into Eq. (22) and then van-ishing the coefficients of zi, one can get some algebraic equa-tions about a0, a1, A and ν, which solving them by Maple,
concludes: • Case 1: a0= 0, a1= b¡µ2∓ |µ|ψ¢ 2aµγ , A = p −µ2+ 2λγ ± |µ|ψ 2a√γ , ν = µ 2− 6λγ ∓ |µ|ψ 2pγ (−µ2+ 2λγ ± |µ|ψ),
where ψ = pµ2− 4λγ. In this case, the exact solutions of Eq. (22) are: Θ(ξ) = b(−µ 2± |µ|ψ) exp£a(ξ + ξ 0) ¤ 2γµ¡−1 + b exp£a(ξ + ξ0) ¤¢ , a > 0, b < 0, Θ(ξ) = b(−µ 2± |µ|ψ) exp£a(ξ + ξ 0) ¤ 2γµ¡1 + b exp£a(ξ + ξ0) ¤¢ , a < 0, b > 0,
and using the substitution in (18) we get the final solutions:
u(x, t) =b(−µ 2± |µ|ψ exp£η(x, t)¤) 2γµ¡−1 + b exp£η(x, t)¤¢ , a > 0, b < 0, u(x, t) =b(−µ 2± |µ|ψ) exp£η(x, t)¤ 2γµ¡1 + b exp£η(x, t)¤¢ , a < 0, b > 0, where η(x, t) = 2x p γ(−µ2+ 2γλ ± |µ|ψ) 4γ +(−µ 2+ 6γλ ± |µ|ψ)tα+ 4ξ 0γaα 4γα . • Case 2: a0=ψ − µ 2γ , a1= b(3µ2− 12λγ − µψ) γa(3ψ − µ) , A = √ 2ψ 2a√−γ, ν = √ 2µ 2√−γ.
In this case, the exact solutions of Eq. (22) are:
Θ(ξ) = −2µ 2+ 2µψ + 6λγ + b¡6λγ − µ2− µψ¢exp£a(ξ + ξ 0) ¤ γ(3ψ − µ)¡−1 + b exp£a(ξ + ξ0) ¤¢ , a > 0, b < 0, Θ(ξ) = 2µ 2+ 2µψ − 6λγ + b¡6λγ − µ2− µψ¢exp£a(ξ + ξ 0) ¤ γ(3ψ − µ)¡1 + b exp£a(ξ + ξ0) ¤¢ , a < 0, b > 0, or equivalently u(x, t) =−2µ 2+ 2µψ + 6λγ + b¡6λγ − µ2− µψ¢exp£η(x, t)¤ γ(3ψ − µ)¡−1 + b exp£η(x, t)¤¢ , a > 0, b < 0, u(x, t) =2µ 2+ 2µψ − 6λγ + b¡6λγ − µ2− µψ¢exp£η(x, t)¤ γ(3ψ − µ)¡1 + b exp£η(x, t)¤¢ , a < 0, b > 0, where η(x, t) = ξ0a +xψ √ −2γ 2γ − µψtα 2γα. • Case 3: a0= ψ − µ 2γ , a1= 2b(µ2− 3λγ − µψ) γa(3ψ − µ) , A = p 2λγ − µ2+ µψ 2a√γ , ν = 6λγ − µ2+ µψ 2pγ (2λγ − µ2+ µψ).
In this case, we can obtain
Θ(ξ) = 2 ¡ −µ2+ 3λγ + µψ¢ γ(3ψ − µ)¡−1 + b exp£a(ξ + ξ0) ¤¢ , a > 0, b < 0, Θ(ξ) = 2 ¡ µ2− 3λγ − µψ¢ γ(3ψ − µ)¡1 + b exp£a(ξ + ξ0) ¤¢ , a < 0, b > 0,
and using the substitution in (18) we have
u(x, t) = 2 ¡ −µ2+ 3λγ + µψ¢ γ(3ψ − µ)¡−1 + b exp£η(x, t)¤¢ , a > 0, b < 0, u(x, t) = 2 ¡ µ2− 3λγ − µψ¢ γ(3ψ − µ)¡1 + b exp£η(x, t)¤¢ , a < 0, b > 0, (24) where η(x, t) = ξ0a +x p −µ2+ µψ + 2γλ 2√γ −−µ 2+ µψ + 6γλ 4γα t α.
Also, in the use of Riccati equation, substituting (23) along with (21) into Eq. (22) and then vanishing the coeffi-cients of zi, we can obtain some algebraic equations about
a0, a1, A and ν, that solving them by Computer algebra tech-nique , concludes: • Case 1: a0= − µ 2γ, a1= ± ibψ 2γ√ab, A = ± √ 2ψ 4√γab, ν = i√2µ 2√γ.
In this case, the exact solutions of Eq. (22) are:
Θ(ξ) = −µ ± iψ tan £√ abξ + ξ0 ¤ 2γ , ab > 0, Θ(ξ) = −µ ± ψ tanh £√ −abξ − ² ln(ξ0) 2 ¤ 2γ , ab < 0,
and using the substitution in (18) we get the following final solutions: u(x, t) = − 1 2γ µ µ + iψ tan · −√2γψxα ∓ 4γξ0α + iµψtα 4γα ¸¶ , when ab > 0 and u(x, t) = − 1 2γ µ µ + ψ tanh · −i√2γψxα ± 2γ² ln(ξ0)α − µψtα 4γα ¸¶ , when ab < 0. • Case 2: a0= − µ 2γ, a1= ∓ ibψ 2γ√ab, A = ± √ 2ψ 4√γab, ν = − i√2µ 2√γ ,
Exact solutions of Eq. (22) extracted from this case are:
Θ(ξ) = −µ ∓ iψ tan £√ abξ + ξ0 ¤ 2γ , ab > 0, Θ(ξ) = −µ ∓ ψ tanh £√ −abξ −² ln(ξ0) 2 ¤ 2γ , ab < 0, or equivalently u(x, t) = − 1 2γ µ µ + iψ tan · √ 2γψxα ± 4γξ0α + iµψtα 4γα ¸¶ , when ab > 0 and u(x, t) = − 1 2γ µ µ + ψ tanh · i√2γψxα ∓ 2γ² ln(ξ0)α − µψtα 4γα ¸¶ , when ab < 0.
5.
Conclusion
In this study, the Lie group analysis method was successfully applied to investigate the reduction and symmetry properties of the TFKPP equation. Moreover, we have arrived to some exact solutions of the conformable TFKPP equation, thanks to the application of simplest equation method. The results of this study undoubtedly offer helpful information about the TFKPP equation.
Acknowledgments
The authors would like to thank the referees for the helpful suggestions.
i. In this paper, the case k = 2 has been used to find solutions.
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